%% LyX 2.3.7 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{amsmath}
\usepackage{babel}
\begin{document}
\title{The Cosine of 3 Degrees}
\author{Amit Yaron}
\date{July 1, 2021}
\maketitle
\begin{abstract}
Wow, I saw today on YouTube that there's a video clip explaining how
to calculate the cosine of 3 degrees. The clip is 26 minutes long
with ads. Now, there is a famous exercise, which is to calculte the
sine of 18 degrees. Calculating the sine and cosine of 15 degrees
is easier thanks to the famoues sines and cosines of 30, 45 and 60
degrees. Let's get started.
\end{abstract}

\paragraph{First let us calculate the sine and cosine of 15 degrees:}

\[
\cos(15^{\circ})=\cos(45^{\circ}-30^{\circ})=\cos(45^{\circ})(\cos(30^{\circ})+\sin(45^{\circ})\sin(30^{\circ})=\frac{\sqrt{2}}{2}(\frac{\sqrt{3}}{2}+\frac{1}{2})=\frac{\sqrt{6}+\sqrt{2}}{4}
\]

and

\[
\sin(15^{\circ})=\sin(45^{\circ}-30^{\circ})=\sin(45^{\circ})\cos(30^{\circ})-\cos(45^{\circ})\sin(30^{\circ})=\frac{\sqrt{2}}{2}(\frac{\sqrt{3}}{2}-\frac{1}{2})=\frac{\sqrt{6}-\sqrt{2}}{4}
\]


\paragraph*{Now, let us calculate the cosine and sine of 18 degrees}

\begin{multline*}
360=5\cdot72=2\cdot72+3\cdot72\Rightarrow\\
\cos(2\cdot72^{\circ})=\cos(3\cdot72^{\circ})
\end{multline*}

Now, let $x=72^{\circ}$

Then use the following identities:

\[
\cos(2x)=\cos^{2}(x)-\sin^{2}(x)=\cos^{2}(1)-[1-\cos^{2}(x)]=2\cos^{2}(x)-1
\]

and

\begin{multline*}
\cos(3x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)=\\
=[2\cos^{2}(x)-1]\cos(x)-2\sin^{2}(x)\cos(x)=\\
=[2\cos^{2}(x)-1-2\sin^{2}(x)]\cos(x)=\\
=[2\cos^{2}(x)-1-2+2\cos^{2}(x)]\cos(x)=\\
=[4\cos^{2}(x)-3]\cos(x)=4\cos^{3}(x)-3\cos(x)
\end{multline*}

From the two identities:

\begin{multline*}
2\cos^{2}(x)-1=4\cos^{3}(x)-3\cos(x)\Rightarrow\\
\Rightarrow4\cos^{3}(x)-2\cos^{2}(x)-3\cos(x)+1=0
\end{multline*}

Let $t=\cos(x)$ :

\begin{multline*}
4t^{3}-2t^{2}-3t+1=0\Rightarrow\\
\Rightarrow4t^{3}-4t^{2}+2t^{2}-2t-t+1=0\Rightarrow\\
\Rightarrow(t-1)(4t^{2}+2t-1)=0
\end{multline*}
 

but $t=\cos(72^{\circ})\neq\cos(0^{\circ})=1$, so let us find other
values using the quadratic equation:

\begin{multline*}
4t^{2}+2t-1=0\Rightarrow\\
\Rightarrow t=\frac{-2\pm\sqrt{2^{2}+4\cdot4\cdot1}}{8}=\frac{-2\pm\sqrt{20}}{8}=\frac{-2\pm2\sqrt{5}}{8}=\frac{-1\pm\sqrt{5}}{4}
\end{multline*}

Because $\cos(72^{\circ})>0$:

\begin{multline*}
\sin(18^{\circ})=\cos(72^{\circ})=\frac{\sqrt{5}-1}{4}
\end{multline*}

The sine is:

\begin{multline*}
\cos(18^{\circ})=\sqrt{1-\sin^{2}(18^{\circ})}=\\
=\sqrt{1-(\frac{\sqrt{5}-1}{4})^{2}}=\\
=\sqrt{1-\frac{6-2\sqrt{5}}{16}}=\\
=\sqrt{\frac{10+2\sqrt{5}}{16}}=\\
=\frac{\sqrt{10+2\sqrt{5}}}{4}\\
\end{multline*}

Now, the requested cosine can be expressed using radicals:

\begin{multline*}
\cos(3^{\circ})=\cos(18^{\circ}-15^{\circ})=\cos(18^{\circ})\cos(15^{\circ})+\sin(18^{\circ})\cos(15^{\circ})=\\
=(\frac{\sqrt{10+2\sqrt{5}}}{4})(\frac{\sqrt{6}+\sqrt{2}}{4})+(\frac{\sqrt{5}-1}{4})(\frac{\sqrt{6}-\sqrt{2}}{4})=\\
=\frac{\sqrt{60+12\sqrt{5}}+\sqrt{20+4\sqrt{5}}+\sqrt{30}-\sqrt{6}-\sqrt{10}+\sqrt{2}}{16}
\end{multline*}

\end{document}
